Geometric Algebra in 2D - Linear Algebra and Cramer's Rule

Рет қаралды 19,055

Mathoma

Күн бұрын

Пікірлер: 52

@thomives4353 2 жыл бұрын

This was AMAZING! Thanks for a great lecture!

@wyboo2019 2 жыл бұрын

18:37 you should have parentheses around the denominator, eg (2+1) e1 e2, because what you have written down, 2 + 1 e1 e2, looks like a multivector with scalar part 2 and bivector part 1

@MultivectorAnalysis 8 жыл бұрын

Another outstanding video! I really like your presentation style. I haven't had time to watch your older videos on quaternions, but judging by the thumbnails, I'm really looking forward to seeing your work with G(3). I love how straightforward and natural determinants are in GA. Defining the determinant of a linear transformation as the proportionality constant of that linear transformation acting on a pseudoscalar gives geometric motivation and intuition for all those determinant properties that many students studying matrix theory find so mysterious. BTW, I had several students ask me about a GA shirt I was wearing yesterday, and they seemed really interested in learning more. I didn't think of it at the time, but I'm going to point them to this series as an excellent place to start their GA exploration.

@Math_oma 8 жыл бұрын

+Nick OkamotoThanks! I always loathed how the determinant is often introduced as a mysterious "number" tied to a square matrix with some computational tricks, crossing out rows and columns. Some recent progress I've seen is 3blue1brown's video series "Essence of Linear Algebra" where the determinant is introduced immediately as a scaling factor on the parallelotope formed by the basis vectors. Then, the tricks used to compute it are seen as subsidiary. You'll probably notice that in those quaternion videos that all I'm doing is reverse engineering some quaternions formulas using more familiar concepts like the dot product and cross product. However, those videos were my attempt to help counter some of the obfuscation surrounding why quaternions work (and my own attempt to better understand why they work).

@MultivectorAnalysis 8 жыл бұрын

Exactly! Then things like the invertibility of a linear transformation being dependent on whether or not the determinant is zero acquire geometric intuition. If the determinant is zero, then at least one dimension of the parallelotope is collapsed, dashing our hopes of finding an inverse transformation. Thanks for pointing me to 3blue1brown's videos; I wasn't aware of that channel. I just skimmed a few of his videos, and they look great!

@Math_oma 8 жыл бұрын

+Nick Okamoto Yeah I was very pleased to see videos that explain linear algebra not from the point of view of slavishly finding solutions of systems of equations but from the linear transformation point of view.

@jongxina3595 3 жыл бұрын

Damn linear algebra feels like such a lie after seeing all this. I was always so annoyed at cross product, determinant, cramers rule, etc; and how they all seemed to come out of nowhere. This field is so elegant and its not even black magic either.

@Fractured_Scholar 2 жыл бұрын

@Mathoma - Around 28:50, shouldn't that say y(b ∧ a)=0? Or -y(a ∧ b)=0? I'm still new at this, so I definitely could be wrong.

@rzezzy1 5 жыл бұрын

Can you recommend a source for practice problems? While these videos are great and they make me feel as though it all makes perfect sense, I'm an active learner and I'm struggling to find resources to help me practice outside of just watching.

@kx4532 3 жыл бұрын

Second that! Maybe with solutions for odd problems.

@sjkebab 10 ай бұрын

Thirded

@Adolphout 9 ай бұрын

Geometric Algebra for Physicists by Doran and Lasenby. Clifford Algebras and Spinors by Lounesto.

@dliessmgg 5 жыл бұрын

Hi! Do you have a video on the Interior Product? (i.e. the product that _decreases_ the rank/grade of a multivector, kinda like an opposite of the wedge product.) I'm trying to figure out some abstract music theory that uses these things and I can't find good, understandable descriptions of it.

@Buzzard-wq1bw Жыл бұрын

Wouldn't that be the Dot Product? Or are you looking for the Regressive/Anti-Wedge Product?

@johnhare8208 5 жыл бұрын

Since the planes are amorphous you can take the limit as you elongate it and manipulation the sheers as the vector of the unit area sphere

@demr04 3 жыл бұрын

Thanks for your videos on this topic. I have one question: why you can divide by a bivector? In your previous videos you define the inverse vector, but i don't understand how you do de first. I was think in taking the absolute of bivector because x and y are scalar. And I think that make sense.

@daanvanijcken4288 3 жыл бұрын

I have exactly this question. We haven't even defined the norm operator for bivectors so even u/ ||u||^2 = u^-1 cannot yet be done. In addition, how would we divide if our result consisted of the sum of a vector and a bivector?

@demr04 3 жыл бұрын

@@daanvanijcken4288 i finish this playlist and also watched another videos of GA, and didnt see any real division of multivector, i.e that is really computable.

@davidoobarber 3 жыл бұрын

@@demr04 I'm no expert, but I think if you look at x a^b = c^b and express a^b in terms of a basis e1, e2, then any wedge product will be some scalar s(a,b) times e1^e2. Hence x a^b = c^b can be expressed as x s(a,b)e1^e2 = s(c,b)e1^e2. For this equality to hold we must therefore have x=s(c,b)/s(a,b), which defines what is meant by division in this case.

@demr04 3 жыл бұрын

@@davidoobarber Thanks. When I watched the video, I forget that Clifford in R2, bivector are just 1-d objects, so dividing is define because you are doing algebra in one set of numbers.

@person1082 2 жыл бұрын

@@demr04 for a bivector, a^2=-|a|^2, so 1/a^2=-1/|a|^2, multiply by a to get 1/a=-a/|a|^2, the - sign is there for 2, 3, 6, 7,… dimensions

@lt4376 3 жыл бұрын

18:31 the denominator (2+1)e1^e2 looked like 2+1e1^e2 at first…. My fault running through the video… but it is very interesting because the isn’t wedge or outer product the same as the matrix product of two unit vectors resulting in a matrix? I say that because the division of a wedge product would appear to be the same as inverting that matrix product… so I wouldn’t have to find an inverse anymore but just look to the geometric algebra viewpoint to find my resultant..

@sang1s160 5 ай бұрын

The power of a scalar + bivector is magical

@CephalopodParty 8 жыл бұрын

Hey I watched your set theory videos forever ago and really liked them, recently I've been struggling to understand the tensor algebra and its quotient spaces and it looked like this series would help me with that. It seems that what you call the geometric product is the tensor product and what you call the geometric algebra is the tensor algebra of a real vector space. If this is the case I must thank you for explaining it in such an elegant way-I was pretty lost before. I also want to know what you would recomend for learning about the symmetric, exterior, and clifford algebras.

@CephalopodParty 8 жыл бұрын

I actually meant Euclidean space by real vector space, also I realized that the tensor algebra is the infinite direct sum of a vector space tensored with itself, but what you're talking about seems to be a finite direct sum from 0 to the dimendion of the space.

@kucckumelon2837 Жыл бұрын

How has your learning progress been?

@CephalopodParty Жыл бұрын

@kucckmelon i asked this in high school and im now a grad student so i guess pretty great XD

@kucckumelon2837 Жыл бұрын

@CephalopodParty oh great! May i ask which course are you taking for grad school? I'm a first year physics student and i just found your 7 year old comment so i figured I'll ask.

@CephalopodParty Жыл бұрын

@@kucckumelon2837 im specializing in discrete math, currently some combinatorics courses + a number theory course

@patryk_49 3 жыл бұрын

I just realised that the complex numbers, quternions and cross products are a bloated APIs for geometric algebra.

@Matthew-by2xx 6 ай бұрын

I mean not really. Complex numbers are much simpler than Geometric algebra, but they are less general.

@IntegralMoon 7 жыл бұрын

Awesome video again! This one I really liked :) I tried continuing this to find the 3d version of cramers rule. Assuming a system was xa + yb + zc = d, I'd imagine it should just be: x = (d ^ b ^ c) / (a^ b ^ c) [taking a few liberties with notation] Which got me thinking about something interesting. To prove that, I'd have to assume that the wedge product is associative. I know that can be proved, but its so strange to me that the wedge product is associative given something like the cross product is not. I thought the cross product and the wedge product were basically the same before thinking about this, now I can see there are some differences. Am I mistaken about any of this perhaps?

@Math_oma 7 жыл бұрын

+IntegralMoon Looks right to me. Yes, the wedge product is associative although I don't think I've proved that in this series yet (an exercise for the viewer, then). It's quite reasonable to expect this geometrically if you think of the wedge product in the extensive manner, that is, extending one object along another to form a new one. The cross product is unnatural in that it only makes sense to talk about cross products in R^3 whereas the wedge product is good in all dimensions and the cross product is not associative, as you know. In this series, the cross product will be defined in terms of the wedge product, namely its dual.

@IntegralMoon 7 жыл бұрын

I appreciate you taking the time to respond to me on this :) What I find confusing though is the algebra. Does defining the cross product as the dual of the wedge product in 3d destroy the associativity of the cross product. My understanding is that the cross product vector is perpendicular to the bivector formed by its two components; what I'm failing to see is why the cross product isn't associative if the dual product is. The only way I can make sense of this is, if I had three vectors a, b and c, I could get: (a ^ b ) ^ c = a ^ ( b ^ c) But the act of taking a ^ b ^ c produces a 3d bivector, for which the dual would be a 0d bivector (in R^3) if I'm not mistaken. Of course the cross product a x b x c produces a new vector. How could you compute that same vector using a wedge product?

@Math_oma 7 жыл бұрын

+IntegralMoon The cross product isn't associate to begin with so its associativity cannot be destroyed. This will be covered in a later video but as a hint, play around with multiplying the pseudoscalar e1e2e3 by each of the three basis bivectors, e1e2, e2e3, and e3e1 a look at the vector that results from this multiplication and its relation to the input bivector. From here, you should be able to define the cross product in terms of the wedge product and pseudoscalar. This definition makes the cross product merely derivative, not fundamental, in geometric algebra. Remember that wedging three vectors produces a trivector, not a bivector - that is, a grade-3 element of G(3). Yes, that's right, the dual of the grade-3 trivector will be grade-0 but this is simply a scalar, not a 0d bivector (?). Remember the graded hierarchy of basis elements in G(3).

@IntegralMoon 7 жыл бұрын

Sorry, I have my terminology a little wrong there. Thanks for correcting that :) I'll give it a go. Thanks again for this awesome resource!

@miroslavjosipovic5014 5 жыл бұрын

@@IntegralMoon The cross product in geometric algebra is different from the standard cross product, just try to apply the space inversion. In fact, this is very important.

@henrmota 4 жыл бұрын

So elegant wow. I'm surprised by geometric algebra.

@cristian-bull 4 жыл бұрын

I can´t stop thinking of a third vector popping out of the screen when you draw the counterclockwise rotation

@skyfall-t8p 3 жыл бұрын

Alternative title for the series: 3b1b's Essence of Linear Algebra, formalized

@angeldude101 2 жыл бұрын

There is also a geometric algebra picture for the row interpretation of the system. It first requires the expression equal 0, so move the constant to the other side and then write out a vector with the corresponding coefficients. Since the constant term is non-zero, you'd need to use PGA in this case which adds a basis vector e₀ with e₀² = 0. This gives v = e₁ - e₂ - e₀ and u = e₁ + 2e₂. Now just wedge the two together for v ∧ u = 2e₁₂ - e₂₁ - e₀₁ - 2e₀₂ = 3e₁₂ + 2e₂₀ - e₀₁. Since e₀² = 0, |v ∧ u| = 3. Normalize the result with that in mind for ⅔e₂₀ - ⅓e₀₁ + e₁₂, which is exactly the point (⅔, -⅓) where the two lines intersect. * Small caveat, I'm using e₂₀ and e₀₁ due to the dual space nature of this picture. e₁ * e₂₀ = e₂ * e₀₁ = e₀₁₂. In some ways, the resulting bivector represents the system of equations itself more than the actual point. It is also the (point) axis of rotation when composing the two vectors as mirrors.

@Fractured_Scholar 2 жыл бұрын

Could you add 0e₀ anywhere it belongs in your examples to make them easier to see? :)

@angeldude101 2 жыл бұрын

@@Fractured_Scholar Uh... sure? That would be akin to writing 1x + 2y + 0 = 0 instead of x + 2y = 0, but doing so would make 1e₁ + 2e₂ + 0e₀ = u. Interestingly, nothing else in the comment seems to have omitted 0 terms.

@Fractured_Scholar 2 жыл бұрын

@@angeldude101 - Thanks. Now that it isn't reduced, we can read the values of the vector straight off the equation :)

@angeldude101 2 жыл бұрын

@@Fractured_Scholar GA generally doesn't write vectors as a list of numbers like traditional vector algebra and only writes them as a linear combination of basis vectors. When doing that, all that really matters would probably be lining up the components so that they align. Also, any GA problem can be interpreted as being within the infinite dimensional Universal Geometric Algebra, which would naturally have infinite 0 components, so instead we only write down the components relevant to us. The difference between PGA and VGA could be seen as just whether or not planes can have a non-zero constant term, but of course they can also have other terms, including quadratic ones or even more complex ones. I could just as easily include "+ 0x² + 0y²" and it would mean the exact same thing, only in CGA rather than PGA.